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Let . Does there exist a constant such that, for all primes , every residue modulo is the sum of at most many elements of
where denotes the inverse of modulo ?
Source: erdosproblems.com/1180
An accepted solution exists. The statement is true.
Proved, in the site's label. Shparlinski's Theorem 3 (Arch. Math. 78 (2002), 445--448, refereed), an accepted full claim on its page, gave the first affirmative answer: for every , every sufficiently large prime and every integer , pairwise distinct with , the site's for , which the authored small-prime remark below extends to every prime with repetition allowed. Also first-hand: Croot's Theorem 2 with (Integers 4 (2004), Paper A20, refereed; the journal's text agrees with arXiv v2), on its page, for every an such that every residue modulo every prime is a sum of inverses of integers in , read with at most summands, as the question asks and the proof's small-prime step gives (as printed, with exactly , the primes with fail), which extends to all by the monotonicity remark; and Glibichuk's Theorem 3 (Mat. Zametki 79 (2006), 384--395, refereed; in Russian), on its page, pairwise distinct summands for all sufficiently large , the site's , which an authored one-line remark below extends to the remaining finitely many primes with repetition allowed. The three pages are accepted full claims, and the frontmatter standing is derived from them. The trivial lower bound (the site's remark) follows from counting; the true order of between and is open.